Question

Calculate the bond dissociation energy (in kJ/mol) for the O-H bond in water (H2O) given that the enthalpy change for the reaction H2O (g) → H (g) + OH(g) is +495 kJ/mol and the enthalpy change for the reaction H2(g) + 1/2O2(g) → H2O(g) is -241.8 kJ/mol.

Answer 1

To calculate the bond dissociation energy for the O-H bond in water, we need to find the enthalpy change for the reaction H2O g   2H g  + 1/2O2 g . We can do this by using the given reactions and Hess's Law.First, let's write down the given reactions:1  H2O g   H g  + OH g  ; H1 = +495 kJ/mol2  H2 g  + 1/2O2 g   H2O g  ; H2 = -241.8 kJ/molNow, we need to manipulate these reactions to obtain the desired reaction:3  H2O g   2H g  + 1/2O2 g To do this, we can reverse reaction 2 and add it to reaction 1:-1 * [H2 g  + 1/2O2 g   H2O g ] ; -1 * H2 = +241.8 kJ/molH2O g   H2 g  + 1/2O2 g  ; H3 = +241.8 kJ/molNow, add reaction 1 and the modified reaction 3:H2O g   H g  + OH g  ; H1 = +495 kJ/molH2O g   H2 g  + 1/2O2 g  ; H3 = +241.8 kJ/mol----------------------------------------------2H2O g   H g  + OH g  + H2 g  + 1/2O2 g Now, we can cancel out the H2 g  and 1/2O2 g  on the right side of the equation:2H2O g   H g  + OH g  + H g Combine the H g  on the right side:2H2O g   2H g  + OH g Now, divide the entire equation by 2:H2O g   H g  + 1/2OH g Now, we can find the enthalpy change for this reaction by adding the enthalpy changes of reaction 1 and the modified reaction 3:H = H1 + H3 = +495 kJ/mol + +241.8 kJ/mol = +736.8 kJ/molThe bond dissociation energy for the O-H bond in water is 736.8 kJ/mol.